3.919 \(\int \frac{(A+B x) \sqrt{a+b x+c x^2}}{x^5} \, dx\)

Optimal. Leaf size=172 \[ -\frac{(2 a+b x) \sqrt{a+b x+c x^2} \left (-4 a A c-8 a b B+5 A b^2\right )}{64 a^3 x^2}+\frac{\left (b^2-4 a c\right ) \left (-4 a A c-8 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{128 a^{7/2}}+\frac{(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4} \]

[Out]

-((5*A*b^2 - 8*a*b*B - 4*a*A*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(64*a^3*x^2) - (A*(a + b*x + c*x^2)^(3/2))/
(4*a*x^4) + ((5*A*b - 8*a*B)*(a + b*x + c*x^2)^(3/2))/(24*a^2*x^3) + ((b^2 - 4*a*c)*(5*A*b^2 - 8*a*b*B - 4*a*A
*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(128*a^(7/2))

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Rubi [A]  time = 0.142699, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {834, 806, 720, 724, 206} \[ -\frac{(2 a+b x) \sqrt{a+b x+c x^2} \left (-4 a A c-8 a b B+5 A b^2\right )}{64 a^3 x^2}+\frac{\left (b^2-4 a c\right ) \left (-4 a A c-8 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{128 a^{7/2}}+\frac{(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^5,x]

[Out]

-((5*A*b^2 - 8*a*b*B - 4*a*A*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(64*a^3*x^2) - (A*(a + b*x + c*x^2)^(3/2))/
(4*a*x^4) + ((5*A*b - 8*a*B)*(a + b*x + c*x^2)^(3/2))/(24*a^2*x^3) + ((b^2 - 4*a*c)*(5*A*b^2 - 8*a*b*B - 4*a*A
*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(128*a^(7/2))

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a+b x+c x^2}}{x^5} \, dx &=-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}-\frac{\int \frac{\left (\frac{1}{2} (5 A b-8 a B)+A c x\right ) \sqrt{a+b x+c x^2}}{x^4} \, dx}{4 a}\\ &=-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac{(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}+\frac{\left (5 A b^2-8 a b B-4 a A c\right ) \int \frac{\sqrt{a+b x+c x^2}}{x^3} \, dx}{16 a^2}\\ &=-\frac{\left (5 A b^2-8 a b B-4 a A c\right ) (2 a+b x) \sqrt{a+b x+c x^2}}{64 a^3 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac{(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}-\frac{\left (\left (b^2-4 a c\right ) \left (5 A b^2-8 a b B-4 a A c\right )\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{128 a^3}\\ &=-\frac{\left (5 A b^2-8 a b B-4 a A c\right ) (2 a+b x) \sqrt{a+b x+c x^2}}{64 a^3 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac{(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}+\frac{\left (\left (b^2-4 a c\right ) \left (5 A b^2-8 a b B-4 a A c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )}{64 a^3}\\ &=-\frac{\left (5 A b^2-8 a b B-4 a A c\right ) (2 a+b x) \sqrt{a+b x+c x^2}}{64 a^3 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac{(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}+\frac{\left (b^2-4 a c\right ) \left (5 A b^2-8 a b B-4 a A c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{128 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.188942, size = 153, normalized size = 0.89 \[ \frac{\frac{3 \left (-4 a A c-8 a b B+5 A b^2\right ) \left (x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{a} (2 a+b x) \sqrt{a+x (b+c x)}\right )}{16 a^{3/2} x^2}+\frac{(5 A b-8 a B) (a+x (b+c x))^{3/2}}{x^3}-\frac{6 a A (a+x (b+c x))^{3/2}}{x^4}}{24 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^5,x]

[Out]

((-6*a*A*(a + x*(b + c*x))^(3/2))/x^4 + ((5*A*b - 8*a*B)*(a + x*(b + c*x))^(3/2))/x^3 + (3*(5*A*b^2 - 8*a*b*B
- 4*a*A*c)*(-2*Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*x^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sq
rt[a + x*(b + c*x)])]))/(16*a^(3/2)*x^2))/(24*a^2)

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Maple [B]  time = 0.01, size = 569, normalized size = 3.3 \begin{align*} -{\frac{B}{3\,a{x}^{3}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{bB}{4\,{a}^{2}{x}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}B}{8\,{a}^{3}x} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{3}B}{8\,{a}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{{b}^{3}B}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{B{b}^{2}cx}{8\,{a}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{Bcb}{4\,{a}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{Bcb}{4}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{A}{4\,a{x}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,Ab}{24\,{a}^{2}{x}^{3}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,A{b}^{2}}{32\,{a}^{3}{x}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{b}^{3}}{64\,{a}^{4}x} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,A{b}^{4}}{64\,{a}^{4}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,A{b}^{4}}{128}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{7}{2}}}}-{\frac{5\,A{b}^{3}cx}{64\,{a}^{4}}\sqrt{c{x}^{2}+bx+a}}+{\frac{7\,A{b}^{2}c}{32\,{a}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,A{b}^{2}c}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{Ac}{8\,{a}^{2}{x}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{Abc}{16\,{a}^{3}x} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ab{c}^{2}x}{16\,{a}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{A{c}^{2}}{8\,{a}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{A{c}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^5,x)

[Out]

-1/3*B/a/x^3*(c*x^2+b*x+a)^(3/2)+1/4*B/a^2*b/x^2*(c*x^2+b*x+a)^(3/2)-1/8*B/a^3*b^2/x*(c*x^2+b*x+a)^(3/2)+1/8*B
/a^3*b^3*(c*x^2+b*x+a)^(1/2)-1/16*B/a^(5/2)*b^3*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/8*B/a^3*b^2*c*
(c*x^2+b*x+a)^(1/2)*x-1/4*B/a^2*b*c*(c*x^2+b*x+a)^(1/2)+1/4*B/a^(3/2)*b*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^
(1/2))/x)-1/4*A*(c*x^2+b*x+a)^(3/2)/a/x^4+5/24*A/a^2*b/x^3*(c*x^2+b*x+a)^(3/2)-5/32*A/a^3*b^2/x^2*(c*x^2+b*x+a
)^(3/2)+5/64*A/a^4*b^3/x*(c*x^2+b*x+a)^(3/2)-5/64*A/a^4*b^4*(c*x^2+b*x+a)^(1/2)+5/128*A/a^(7/2)*b^4*ln((2*a+b*
x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-5/64*A/a^4*b^3*c*(c*x^2+b*x+a)^(1/2)*x+7/32*A/a^3*b^2*c*(c*x^2+b*x+a)^(1/2
)-3/16*A/a^(5/2)*b^2*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/8*A/a^2*c/x^2*(c*x^2+b*x+a)^(3/2)-1/16*
A/a^3*c*b/x*(c*x^2+b*x+a)^(3/2)+1/16*A/a^3*c^2*b*(c*x^2+b*x+a)^(1/2)*x-1/8*A/a^2*c^2*(c*x^2+b*x+a)^(1/2)+1/8*A
/a^(3/2)*c^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.0235, size = 981, normalized size = 5.7 \begin{align*} \left [-\frac{3 \,{\left (8 \, B a b^{3} - 5 \, A b^{4} - 16 \, A a^{2} c^{2} - 8 \,{\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} \sqrt{a} x^{4} \log \left (-\frac{8 \, a b x +{\left (b^{2} + 4 \, a c\right )} x^{2} + 4 \, \sqrt{c x^{2} + b x + a}{\left (b x + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{2}}\right ) + 4 \,{\left (48 \, A a^{4} -{\left (24 \, B a^{2} b^{2} - 15 \, A a b^{3} - 4 \,{\left (16 \, B a^{3} - 13 \, A a^{2} b\right )} c\right )} x^{3} + 2 \,{\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2} + 12 \, A a^{3} c\right )} x^{2} + 8 \,{\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt{c x^{2} + b x + a}}{768 \, a^{4} x^{4}}, \frac{3 \,{\left (8 \, B a b^{3} - 5 \, A b^{4} - 16 \, A a^{2} c^{2} - 8 \,{\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (b x + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{2} + a b x + a^{2}\right )}}\right ) - 2 \,{\left (48 \, A a^{4} -{\left (24 \, B a^{2} b^{2} - 15 \, A a b^{3} - 4 \,{\left (16 \, B a^{3} - 13 \, A a^{2} b\right )} c\right )} x^{3} + 2 \,{\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2} + 12 \, A a^{3} c\right )} x^{2} + 8 \,{\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt{c x^{2} + b x + a}}{384 \, a^{4} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^5,x, algorithm="fricas")

[Out]

[-1/768*(3*(8*B*a*b^3 - 5*A*b^4 - 16*A*a^2*c^2 - 8*(4*B*a^2*b - 3*A*a*b^2)*c)*sqrt(a)*x^4*log(-(8*a*b*x + (b^2
+ 4*a*c)*x^2 + 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 4*(48*A*a^4 - (24*B*a^2*b^2 - 15*A
*a*b^3 - 4*(16*B*a^3 - 13*A*a^2*b)*c)*x^3 + 2*(8*B*a^3*b - 5*A*a^2*b^2 + 12*A*a^3*c)*x^2 + 8*(8*B*a^4 + A*a^3*
b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^4), 1/384*(3*(8*B*a*b^3 - 5*A*b^4 - 16*A*a^2*c^2 - 8*(4*B*a^2*b - 3*A*a*b^
2)*c)*sqrt(-a)*x^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 2*(48*A*a^
4 - (24*B*a^2*b^2 - 15*A*a*b^3 - 4*(16*B*a^3 - 13*A*a^2*b)*c)*x^3 + 2*(8*B*a^3*b - 5*A*a^2*b^2 + 12*A*a^3*c)*x
^2 + 8*(8*B*a^4 + A*a^3*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \sqrt{a + b x + c x^{2}}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**(1/2)/x**5,x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**5, x)

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Giac [B]  time = 1.42301, size = 1338, normalized size = 7.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^5,x, algorithm="giac")

[Out]

1/64*(8*B*a*b^3 - 5*A*b^4 - 32*B*a^2*b*c + 24*A*a*b^2*c - 16*A*a^2*c^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))/sqrt(-a))/(sqrt(-a)*a^3) - 1/192*(24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a*b^3 - 15*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))^7*A*b^4 - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^2*b*c + 72*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^7*A*a*b^2*c - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*c^2 - 384*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^6*B*a^3*c^(3/2) - 88*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*b^3 + 55*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^5*A*a*b^4 - 288*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^3*b*c - 264*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^5*A*a^2*b^2*c - 336*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^3*c^2 - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^4*B*a^3*b^2*sqrt(c) + 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^4*c^(3/2) - 1152*(sqrt(c)*x - sqrt(c*
x^2 + b*x + a))^4*A*a^3*b*c^(3/2) + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^3*b^3 - 73*(sqrt(c)*x - sqrt(
c*x^2 + b*x + a))^3*A*a^2*b^4 + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^4*b*c - 648*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^3*A*a^3*b^2*c - 336*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*c^2 + 384*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^2*B*a^4*b^2*sqrt(c) - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^3*b^3*sqrt(c) - 128*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))^2*B*a^5*c^(3/2) - 256*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^4*b*c^(3/2) + 24*
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*b^3 - 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b^4 + 288*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))*B*a^5*b*c - 312*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^4*b^2*c - 48*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))*A*a^5*c^2 + 128*B*a^6*c^(3/2) - 128*A*a^5*b*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^2 - a)^4*a^3)